Activated Sludge Process - Wastewater Operators - page 327

Solution:

145mg/L2.2 MGD8.34lb/MG/mg/L
2

× ×
550mg/L0.65MG8.34lb/MG/mg/L

F/MRatio = = 0.19 lb B OD/lbMLVSS

× ×

Note: If the MLVSS concentration is not available, it can be calculated
if percent volatile matter (%VM) of the mixed liquor suspended solids
(MLSS) is known (see Equation 8.11).

MLVSS = MLSS × %VM (8.11)

Note: The “F” value in the F/M ratio for computing loading to an activated
sludge process can be either BOD or COD. Remember that the reason
for sludge production in the activated sludge process is to convert BOD
to bacteria. One advantage of using COD over BOD for the analysis of
organic load is that COD is more accurate.

 Example 8.6

Problem: The aeration tank contains 2985 mg/L of MLSS. Laboratory
tests indicate the MLSS is 66% volatile matter. What is the MLVSS con-
centration in the aeration tank?

Solution:

MLVSS = 2985 mg/L × 0.66 = 1970 mg/L

8.14.5.1 F/M Ratio Control

Maintaining the F/M ratio within a specified range can be an excel-
lent control method. Although the F/M ratio is affected by adjustment
of the return rates, the most practical method for adjusting the ratio is
through waste rate adjustments:

Increasing the waste rate will …

• Decrease the MLVSS.

• Increase the F/M ratio.

Decreasing the waste rate will …

• Increase the MLVSS.

• Decrease the F/M ratio.

The desired F/M ratio must be established on a plant-by-plant basis.
Comparison of F/M ratios with plant effluent quality is the primary means
for identifying the most effective range for individual plants, when the
range of F/M values that produce the desired effluent quality has been
established.